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Question: If\(\int_{\sin x}^{1}t^{2}\) (f(t)) dt = (1 – sin x), then f\(\left( \frac{1}{\sqrt{3}} \right)\) is...

If∫sin⁡x1t2\int_{\sin x}^{1}t^{2} (f(t)) dt = (1 – sin x), then f(13)\left( \frac{1}{\sqrt{3}} \right) is –

A

13\frac{1}{3}

B

13\frac{1}{\sqrt{3}}

C

3

D

3\sqrt{3}

Answer

3

Explanation

Solution

∫sin⁡x1t2f(t)dt=1–sin⁡x\int_{\sin x}^{1}{t^{2}f(t)dt = 1–\sin x} {given}

Differentiating both sides w.r.t. x using Newton

Leibnitz formula, we have

– sin2 x f (sin x) cos x = – cos x

Ž sin2 x f (sin x) cos x = cos x {cos x ¹ 0}

Ž f(sin x) = 1sin⁡2x\frac{1}{\sin^{2}x}, x ¹ (2n + 1)π2\frac{\pi}{2}

\ f(x) = 1x2\frac{1}{x^{2}}

Now, f(13)\left( \frac{1}{\sqrt{3}} \right) = 3